91Ó°ÊÓ

We need to find the first partial derivatives of the function \(f(x, y) = \left(y^{2}- x^{2}\right) \cdot \mathrm{e}^{-\frac{x^{2} + y^{2}}{2}}\). Let's find the partial derivative with respect to \(x\) and \(y\). $$ f_x(x, y) = \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}\left[\left(y^{2}-x^{2}\right) \cdot \mathrm{e}^{-\frac{x^{2}+y^{2}}{2}}\right] $$ $$ f_y(x, y) = \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left[\left(y^{2}-x^{2}\right) \cdot \mathrm{e}^{-\frac{x^{2}+y^{2}}{2}}\right] $$ By applying the product rule and chain rule, we get: $$ f_x(x, y) = \left(-2x\right) \mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} + \left(y^{2}-x^{2}\right)\left(-x\right)\mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} $$ $$ f_y(x, y) = \left(2y\right) \mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} + \left(y^{2}-x^{2}\right)\left(-y\right)\mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} $$

The critical points occur where both first partial derivatives are equal to zero. So, we need to solve the following equations simultaneously: $$ f_x(x, y) = 0 $$ $$ f_y(x, y) = 0 $$

From the equations in step 2, we get: $$ \left(-2x\right) \mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} + \left(y^{2}-x^{2}\right)\left(-x\right)\mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} = 0 $$ $$ \left(2y\right) \mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} + \left(y^{2}-x^{2}\right)\left(-y\right)\mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} = 0 $$ From both equations, we can factor out the term \(\mathrm{e}^{-\frac{x^{2}+y^{2}}{2}}\), which is never equal to zero. After doing that, we get: $$ -2x - x(y^2 - x^2) = 0 $$ $$ 2y - y(y^2 - x^2) = 0 $$ The critical points occur at \((x, y) = (0, 0)\).

We need to find the second partial derivatives to determine the nature of these critical points: $$ f_{xx}(x, y) = \frac{\partial^2 f}{\partial x^2} $$ $$ f_{yy}(x, y) = \frac{\partial^2 f}{\partial y^2} $$ $$ f_{xy}(x, y) = \frac{\partial^2 f}{\partial x\partial y} $$ After computing, we obtain: $$ f_{xx}(x, y) = \frac{6x^2 - y^2}{\mathrm{e}^{\frac{x^2+y^2}{2}}} $$ $$ f_{yy}(x, y) = \frac{6y^2 - x^2}{\mathrm{e}^{\frac{x^2+y^2}{2}}} $$ $$ f_{xy}(x, y) = f_{yx}(x, y) = -2xy \mathrm{e}^{-\frac{x^{2}+y^{2}}{2}} $$ We need to evaluate the determinant of the Hessian matrix: $$ D(x, y) = f_{xx}(x, y) f_{yy}(x, y) - f_{xy}(x, y) f_{yx}(x, y) $$ At the critical point \((x, y) = (0, 0)\), we get: $$ D(0, 0) = (0)(0) - (-2(0)(0))^2 = 0 $$ Since \(D(0, 0) = 0\), we cannot determine the nature of this critical point using the second derivative test alone. We can try analyzing the original function and use the fact that it's symmetric.

Examining the original function \(f(x, y) = \left(y^{2}- x^{2}\right) \cdot \mathrm{e}^{-\frac{x^{2} + y^{2}}{2}}\), notice that it is symmetric with respect to the \(y=x\) and \(y=-x\) lines. Given this symmetry, it is clear that the critical point at \((0,0)\) is a saddle point. Thus, the function \(f(x, y)\) has a saddle point at \((0, 0)\).